- Thread starter Jason
- Start date

We see this is a type I region, so for a double iterated integral, we will want to integrate with respect to \(y\) first (inner) and then with respect to \(x\) (outer). To determine the outer limits, we can equate the two functions making up the region boundaries:

\(\displaystyle 2x=3-x^2\)

\(\displaystyle x^2+2x-3=0\)

\(\displaystyle (x+3)(x-1)=0\)

\(\displaystyle x\in\{-3,1\}\)

And so:

\(\displaystyle V=\int_{-3}^{1} x^2\int_{2x}^{3-x^2}\,dy\,dx\)

Can you proceed?

How were the boundaries gotten for the other integral - the one without \(\displaystyle 1\) and \(\displaystyle -3\) limits of integration?

OH wait (post edited). I see now! But how did you determine which limit of integration was on the bottom vs which one is on the top?

OH wait (post edited). I see now! But how did you determine which limit of integration was on the bottom vs which one is on the top?

Last edited:

\(\displaystyle \int_{-3}^{1}\,x^{2} \,dy\)

\(\displaystyle \dfrac{x^{3}}{3}\) - evaluated at \(\displaystyle -3\) and \(\displaystyle 1\)

\(\displaystyle [\dfrac{(1)^{3}}{3}] - [\dfrac{(-3)^{3}}{3}] = \dfrac{28}{3}\)

Cannot be done further as a double integral. There aren't any variables left for the 2nd part.

\(\displaystyle \dfrac{x^{3}}{3}\) - evaluated at \(\displaystyle -3\) and \(\displaystyle 1\)

\(\displaystyle [\dfrac{(1)^{3}}{3}] - [\dfrac{(-3)^{3}}{3}] = \dfrac{28}{3}\)

Cannot be done further as a double integral. There aren't any variables left for the 2nd part.

Last edited:

It would be the one I came up with where - once you get the next integral, there's nothing to work with. (guess)Suppose in the double integral, I wished to reverse the order of integration...what issue would I face?

If we reverse the order of integration in the double integral I gave, we would be treating the bounded area as type II, which means we are using horizontal rather than vertical strips. Using horizontal strips, and looking at the bounded region, do you see that the boundary on the right changes from one function to another?It would be the one I came up with where - once you get the next integral, there's nothing to work with. (guess)

So we would certainly need to know what order of integration to use before we attempt the problem. How could we find out?If we reverse the order of integration in the double integral I gave, we would be treating the bounded area as type II, which means we are using horizontal rather than vertical strips. Using horizontal strips, and looking at the bounded region, do you see that the boundary on the right changes from one function to another?

We can evaluate the integral in either order, but I chose the order I did for simplicity. If we treat the area as type II, then we will need two separate integrals. Sometime though, we find that an multiple iterated integral evaluated in one order results in an anti-derivative that is not expressible in elementary functions, but if we reverse or change the order of integration, we don't run into that as an issue. So, it becomes useful to both look at which type is going to be easier, and to be able to switch the order in cases where we need to to be able to actually evaluate the integral in elementary terms.So we would certainly need to know what order of integration to use before we attempt the problem. How could we find out?

That's why I plotted the bounded area first, so that I could see which order made the most sense as a way to begin.

I don't really get what you mean. But I know doing it the way I tried at first caused it to disappear - making it impossible to solve.We can evaluate the integral in either order, but I chose the order I did for simplicity. If we treat the area as type II, then we will need two separate integrals. Sometime though, we find that an multiple iterated integral evaluated in one order results in an anti-derivative that is not expressible in elementary functions, but if we reverse or change the order of integration, we don't run into that as an issue. So, it becomes useful to both look at which type is going to be easier, and to be able to switch the order in cases where we need to to be able to actually evaluate the integral in elementary terms.

That's why I plotted the bounded area first, so that I could see which order made the most sense as a way to begin.